q21.2.5\nwrite the nuclide notation, includin...
q21.2.5\nwrite the nuclide notation, including charge if applicable, for atoms with the following characteristics:\na. 25 protons, 20 neutrons, 24 electrons\nb. 45 protons, 24 neutrons, 43 electrons\nc. 53 protons, 89 neutrons, 54 electrons\nd. 97 protons, 146 neutrons, 97 electrons
Answer
# Explanation:
## Step1: Recall the nuclide notation formula
The nuclide notation is \(_{Z}^{A}X^{n+}\) or \(_{Z}^{A}X^{n-}\), where \(Z\) is the atomic number (number of protons), \(A\) is the mass number (\(A = \text{number of protons}+\text{number of neutrons}\)), \(X\) is the element symbol, and \(n\) is the charge (\(n=\text{number of protons}-\text{number of electrons}\))
## Step2: Find element symbol for each case
- **Case (a)**:
- \(Z = 25\) (protons). The element with \(Z = 25\) is manganese (\(Mn\))
- \(A=25 + 20=45\)
- \(n=25-24 = 1+\)
- Nuclide notation: \(_{25}^{45}Mn^{1+}\)
- **Case (b)**:
- \(Z = 45\) (protons). The element with \(Z = 45\) is rhodium (\(Rh\))
- \(A=45 + 24=69\)
- \(n=45-43 = 2+\)
- Nuclide notation: \(_{45}^{69}Rh^{2+}\)
- **Case (c)**:
- \(Z = 53\) (protons). The element with \(Z = 53\) is iodine (\(I\))
- \(A=53 + 89=142\)
- \(n=53-54=- 1\)
- Nuclide notation: \(_{53}^{142}I^{-}\)
- **Case (d)**:
- \(Z = 97\) (protons). The element with \(Z = 97\) is berkelium (\(Bk\))
- \(A=97+146 = 243\)
- \(n=97 - 97=0\)
- Nuclide notation: \(_{97}^{243}Bk\)
# Answer:
a. \(_{25}^{45}Mn^{1+}\)
b. \(_{45}^{69}Rh^{2+}\)
c. \(_{53}^{142}I^{-}\)
d. \(_{97}^{243}Bk\)